How do you solve using the quadratic formula: $x^2+x+1=0$ ?

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Answer 1 · 2015-11-20T00:14:53

See explanation...

$x^2+x+1$ is of the form $ax^2+bx+c$ with $a = b = c = 1$

This has zeros given by the quadratic formula:

$x = (-b+-sqrt(b^2-4ac))/(2a) = (-1+-sqrt(1^2-(4xx1xx1)))/(2xx1)$

$=(-1+-sqrt(1-4))/2 = (-1+-sqrt(-3))/2 = (-1+-sqrt(3)i)/2$

$= -1/2+-sqrt(3)/2i$

The number $omega = -1/2+sqrt(3)/2i = cos((2pi)/3)+sin((2pi)/3)i$

is called the primitive Complex cube root of unity and is used extensively when solving cubic equations.

$omega^2 = bar(omega) = -1/2-sqrt(3)/2i$

$omega^3 = 1$

How do you solve using the quadratic formula: x^2+x+1=0 x^2+x+1=0?